HDOJ7080Pty hates prime numbers（容斥原理）

Posted 2021-09-09 小哈里

problem

Pty hates prime numbers
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 293 Accepted Submission(s): 87

Problem Description
Pty thinks prime numbers are extremely annoying numbers. Specifically, he hates the first k prime numbers.

As ”Hate me Hate my dog.” , if one prime factor of the integer x is what Pty hates, Pty will also hate x.

Now Pty wants to know how many integers in [1,n] he dosen’t hate.

Formally, Pty wants to know ∑nx=1[∀i∈[1,k],p[i]∤x] (p[i] means the ith prime number) .

Input
The first line contains a single integer T(1 ≤ T ≤ 100000) — the number of test cases

The only line of each test case contains two integers n,k(1 ≤ n ≤ 1018,1 ≤ k ≤ 16).

Output
For each test case, output the number of integers Pty doesn’t hate.

Sample Input
5
10 1
20 2
30 3
40 4
50 5

Sample Output
5
7
8
9
11

Source
2021“MINIEYE杯”中国大学生算法设计超级联赛（10）

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solution

题意：

• 给出n和k，求1-n中有多少个数不是前k个素数中的任意一个的倍数。
• n <=1e18, k <= 16，T<=1e5

思路：

• 不难想到容斥原理，二进制枚举，设枚举的乘积为x，去考虑前k个质数是否选择， 答案为$ans+=n/x\ast (-1{)}^{形成x质数的个数}$。 复杂度2^k，会超时。
• 考虑优化，假设k==8，发现前8个素数的乘积为9699690，注意到如果数x不是前八个素数的倍数的话，那么x%9699690也不是前八个素数的倍数，所以我们可以预处理出1-9699690的答案，对于超过的部分让其==([n/9699690] + n%9699690)进行处理。
  

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL m = 9699690;//2∗3∗5∗7∗11∗13∗17∗19
LL p[20] = {0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};

vector<LL>a[18], b[18];//a,b表示前k个素数中选择偶数/奇数个能形成的数
void dfs(int x, LL y, int z, int k){
    if(x>k){
        if(z==1)a[k].push_back(y);//由偶数个素数组成
        else b[k].push_back(y);
        return ;
    }
    dfs(x+1, y,z,k);       //不使用第x个素数
    dfs(x+1,y*p[x], -z, k);//使用第x个素数
}
LL sum[m+10], as;
void init(){
    for(int k = 1; k <= 16; k++){//暴力前k个素数所能形成的乘积值
        if(k<=8)dfs(1,1,1,k);
        else dfs(9,1,1,k);
    }
    for(int i = 1; i <= 8; i++){ //标记1-m中可以被前8个数整除的数
        for(int j = p[i]; j <= m; j+=p[i]){
            sum[j] = 1;
        }
    }
    for(int i = 1; i <= m; i++){ //维护sum[i]表示1-i中不能被前8个数整除的数的个数
        sum[i] = sum[i-1]+(sum[i]==0);
    }
    as = sum[m];  //as表示1-9699690中不能被前8个数整除的数
}

int main(){
    init();
    int T;  cin>>T;
    while(T--){
        LL n, k;  cin>>n>>k;
        LL ans = 0;
        if(k <= 8){
            for(LL x : a[k])ans += n/x;
            for(LL x : b[k])ans -= n/x;
        }else{
            LL t;
            for(LL x : a[k])t=n/x, ans += (t/m)*as+sum[t%m];
            for(LL x : b[k])t=n/x, ans -= (t/m)*as+sum[t%m];
        }
        cout<<ans<<"\\n";
    }
    return 0;
}